imported>Patrick: Created page with 'For equilibrium, the sum of the three forces must be zero. $F_1 + F_2 + F_3 = 0$ $F_3 = -(F_1 + F_2)$ TOP

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2011-05-12T14:12:38Z

Created page with 'For equilibrium, the sum of the three forces must be zero. <math>F_1 + F_2 + F_3 = 0</math> <math>F_3 = -(F_1 + F_2)</math> TOP We wi…'

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For equilibrium, the sum of the three forces must be zero.

<math>F_1 + F_2 + F_3 = 0</math>
<math>F_3 = -(F_1 + F_2)</math>

[[image:N_APP_EX_20_SOLNa.png|TOP]] 

We will now solve the problem using components method:

[[image:N_APP_EX_20_SOLNa.png|TOP]] 

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_1</math>
| 30 cos(30°) = 8.7 N
| 30 sin(30°) = 5 N
|-
! <math>F_2</math>
| 12.8 cos(-70°) = 4.4 N
| 12.8 sin(-70°) = -12 N
|-
! <math>F_3</math>
| <math>{F_3}_x</math>
| <math>{F_3}_y</math>
|}

We can write the equations for the second Newton Law:

<math>8.7 + 4.4 + {F_3}_x = 0</math>
<math>5 - 12 + {F_3}_y = 0</math>

We find that the force <math>F_3</math> is (-13.1, 7). We can now compute the magnitude and the direction of this force: <math>F_3</math> = 14.9 N and <math>\theta</math> = 152°

Newtons Laws EX 20 - Revision history

imported>Patrick: Created page with 'For equilibrium, the sum of the three forces must be zero. F_1 + F_2 + F_3 = 0 F_3 = -(F_1 + F_2) TOP We wi…'

imported>Patrick: Created page with 'For equilibrium, the sum of the three forces must be zero. $F_1 + F_2 + F_3 = 0$ $F_3 = -(F_1 + F_2)$ TOP

We wi…'